For a normal subgroup $H$ of a group $G$, explain carefully how to make the set of (left) cosets of $H$ into a group.

For a subgroup $H$ of a group $G$, show that the following are equivalent:

(i) $H$ is a normal subgroup of $G$;

(ii) there exist a group $K$ and a homomorphism $\theta: G \rightarrow K$ such that $H$ is the kernel of $\theta$.

Let $G$ be a finite group that has a proper subgroup $H$ of index $n$ (in other words, $|H|=|G| / n)$. Show that if $|G|>n$ ! then $G$ cannot be simple. [Hint: Let $G$ act on the set of left cosets of $H$ by left multiplication.]